5 Steps to a 5 AP Calculus BC 2019

AP Calculus BC Practice Exam 1 401

5. Givenf(0)=1,f′(0)=6,f′′(0)=−4, and
   f′′′(0)= 30.

(A) The third-degree Taylor polynomial forf

aboutx=0is

f(x)≈

f( 0 )

0!

x^0 +

f′( 0 )

1!

x^1 +

f′′( 0 )

2!

x^2

+

f′′′( 0 )

3!

x^3

≈ 1 + 6 x+

− 4

2

x^2 +

30

6

x^3

≈ 1 + 6 x− 2 x^2 + 5 x^3.

To approximatef(0.1):

f(0.1)≈ 1 +6(0.1)−2(0.1)^2 +5(0.1)^3

≈ 1 + 0. 6 −2(0.01)+5(0.001)

≈ 1 + 0. 6 − 0. 02 + 0. 005

≈ 1. 585.

(B) The sixth degree Taylor polynomial for

g(x)= f(x^2 ), aboutx=0, is

g(x)= f

(

x^2

)

≈ 1 + 6

(

x^2

)

− 2

(

x^2

) 2

+ 5

(

x^2

) 3

g(x)≈ 1 + 6 x^2 − 2 x^4 + 5 x^6.

(C) The seventh degree Taylor polynomial for

h(x)=

∫x

0

g(t)dt, aboutx=0, is

h(x)≈

∫x

0

(

1 + 6 t^2 − 2 t^4 + 5 t^6

)

dt

h(x)≈

[

t+

6

3

t^3 −

2

5

t^5 +

5

7

t^7

]x

0

h(x)≈x+ 2 x^3 −

2

5

x^5 +

5

7

x^7.

6. Givenx=2(θ −sinθ) andy=2(1−cosθ):

(A)

dx

dθ

= 2 ( 1 −cosθ)and

dy

dθ

=2 sinθ.

Divide to find

dy

dx

=

2 sinθ

2 ( 1 −cosθ)

=

sinθ

1 −cosθ

.

(B) Atθ=π,x=2(π−sinπ)= 2 π,

y=2(1−cosπ)=4, and

dy

dx

∣∣

∣∣

θ=π

=

sinπ

1 −cosπ

=0.

The tangent line at (2π, 4) is horizontal,

so the equation of the tangent isy=4.

(C) Atθ= 2 π,x=2(2π−sin 2π)= 4 π,

y=2(1−cos 2π)=0, and

dy

dx

∣∣

∣∣

θ= 2 π

=

sin 2π

1 −cos 2π

=

0

0

. Since the

derivative is undefined, the tangent line at

(4π, 0) is vertical, so the equation of the

tangent isx= 4 π.

(D)

L=

∫ 2 π

0

√

[ 2 ( 1 −cosθ)]^2 +[2 sinθ]^2 dθ

=

∫ 2 π

0

√

4(1−2 cosθ+cos^2 θ)+4 sin^2 θdθ

=

∫ 2 π

0

√

4 −8 cosθ+4 cos^2 θ+4 sin^2 θdθ

=

∫ 2 π

0

√

4 −8 cosθ+ 4 dθ

=

∫ 2 π

0

√

8 −8 cosθdθ

= 2

√

2

∫ 2 π

0

√

1 −cosθdθ